On Wed, Feb 16, 2005 at 01:13:53PM +1100, Damian Conway wrote:
> Larry wrote:
> > 0 < $x < 10
> >after all--the problem with rewriting that as
> > 0 < $x and $x < 10
> >is that it should only work as long as the two values of $x remain
> >entangled so that the always refer to the same abstract value.
>
> That's certainly true. But I think the real problem there is in mistakenly
> treating Perl 6 comparators as binary ops (with n-ary sugar), rather than
> as genuine n-ary ops.
>
> That is, a junction must autothread early, and over the entire operation
> (or at least, over the entire equi-precedential part of the operation) in
> which it's used.
Uh oh, I hadn't caught that particular nuance. Is it indeed over the
entire equi-precedential part of the operation, or just over the
chained operators? For example, given
$x = -1 | 10;
$ref.meth1($x).meth2($x)
are the meth1 and meth2 calls considered to be "equi-precedential",
with a result of ...
any( $ref.meth1(-1).meth2(-1), $ref.meth1(10).meth2(10) ) ?
And what of ...
$ref.meth1($x) + $x
are the $x still "tied" to each other even though they're being
used at different levels of precedence? I.e., do I get
any( $ref.meth1(-1) + -1, $ref.meth1(10) + 10)
or
any( any( $ref.meth1(-1) + -1, $ref.meth1(-1) + 10 ),
any( $ref.meth1(10) + -1, $ref.meth1(10) + 10 ) )
?
Pm
